Answer:
a.
Step-by-step explanation:
Assuming the garden is a rectangle, the area is equal to length times width.
So 20x = 760.
Divide both sides by 20.
x = 760/20 = 38 ft
1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
Sin OH Cosin AH Tangent OA search "SOCAHTOA"
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.